Question

A capacitor of capacitance $4 \,\mu F$ is charged to $80\,V$ and another capacitor of capacitance $6\, \mu F$ is charged to $30\, V$. When they are connected together, the energy lost by the $4 \,\mu F$ capacitor is

• A. 7.8 mJ
• B. 4.6 mJ
• C. 3.2 mJ
• D. 2.5 mJ

Answer 1

Answer: (A)

Common potential, $ V=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$
$=\frac{\left(4 \times 10^{-6}\right) \times 80+\left(6 \times 10^{-6}\right) \times 30}{4 \times 10^{-6}+6 \times 10^{-6}}=50 \,V$
$\therefore $ Energy lost by $4 \,\mu F$ capacitor is
$=\frac{1}{2} C_{1} V_{1}^{2}-\frac{1}{2} C_{1} V^{2}=$
$\frac{1}{2} C_{1}\left(V_{1}^{2}-V^{2}\right) $
$=\frac{1}{2} \times\left(4 \times 10^{-6}\right) \times\left\{(80)^{2}-(50)^{2}\right\}$
$=7.8 \times 10^{-3} J =7.8 \,mJ$